Comparing first and second, one may conclude that
$$
x+y+2z = x+2y+z
$$
or equivalently
$$
y = z
$$
Considering other two pairs one may conclude
$$
x = y = z
$$
which is in contradiction with initial assumptions. So the only solutions are those when $x = y = z$.

So either $x=y$ or $x^2 + xy + y^2 + z^2 = 0$. The latter implies that $x=y=z=0$ since the quadratic form is positive definite over $\mathbb{R}$. In either case $x=y$. Other equalities follow similarly from the other minors.

One way to show that the quadratic form is positive definite is to write it as a positive combination of squares: